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How to rearrange

$$y=\sin(x)+2x$$

with $x$ as the subject?

Is this possible?

I'm struggling to work out how I should go about this.

I also tried putting it into wolfram alpha to check how it was worked out but it doesn't return a result.

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As seen here, it is generally impossible to solve for $x$ given a particular $y$, so I doubt you would have much luck solving for $x$ for general $y\in\mathbb Q$. The only closed form solution for rational $y$ occurs at $(0,0)$...

One may, however, approximate the solution using Newton's method:

Let us define $x_n$ as follows:

$$x_{n+1}=x_n-\frac{\sin(x_n)+2x_n-y}{\cos(x_n)+2}$$

Then as $n\to\infty$, $x_n\to x$ is our solution. For example, with $y=1$ and $x_0=0$, we get the following:

$x_1=0-\frac{\sin(0)+0-1}{\cos(0)+2}=\frac12$

$x_2=\frac12-\frac{\sin(1/2)+1-1}{\cos(1/2)+2}=0.335417790$

$x_3=0.335418032$

$x_4=0.335418032$

And thus we have our solution:

$$1\approx\sin(0.335418032)+2(0.335418032)$$

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  • $\begingroup$ It's not true that the only closed form solution occurs at $(0,0)$. For example, for $y = 2+\sin(1)$ the solution is obviously $x=1$. $\endgroup$ Mar 3, 2017 at 23:30
  • $\begingroup$ @RobertIsrael I meant for $y\in\mathbb Q$, sorry. For other $y$ with arguments about what is closed form and what is not, please make comments under the linked question (it would be most useful there) $\endgroup$ Mar 3, 2017 at 23:31

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